Calling a function disrupts the rules - clips

Question

I'm trying to change the design of an expert system to work carried out by my rules.
Topic - processing of different parts on different machines. Naturally, each kind of items processed at different times on different machines.
The system consists of three rules. The first rule - load machines work. The second rule - unloads machines. The third rule - performs the movement of time.
I added in the first rule of a function call, which seeks item with maximum processing time. However, the expert system stopped working. Simply displays "1". That's all.
(defglobal ?*time* = 0)
(deftemplate details
(field det (type SYMBOL))
(field oper (type INTEGER))
(field count (type INTEGER))
)
(deftemplate route
(field det (type SYMBOL))
(field oper (type INTEGER))
(field machine (type INTEGER))
(field time (type INTEGER))
)
(deftemplate machine
(field num (type INTEGER))
(field count (type INTEGER)(default 0))
(field det (type SYMBOL)(default A))
(field oper (type INTEGER)(default 0))
(field time (type INTEGER)(default 0))
)
(deffacts details
(details (det A) (oper 0) (count 100))
(details (det B) (oper 0) (count 150))
(details (det C) (oper 0) (count 200))
(details (det D) (oper 0) (count 300))
(details (det E) (oper 0) (count 200))
(details (det A) (oper 1) (count 0))
(details (det B) (oper 1) (count 0))
(details (det C) (oper 1) (count 0))
(details (det D) (oper 1) (count 0))
(details (det E) (oper 1) (count 0))
....
)
(deffacts route
(route (det A) (oper 1) (machine 1) (time 10))
(route (det A) (oper 2) (machine 2) (time 5))
(route (det A) (oper 2) (machine 2) (time 2))
(route (det A) (oper 3) (machine 3) (time 4))
(route (det A) (oper 3) (machine 4) (time 3))
(route (det A) (oper 4) (machine 4) (time 8))
(route (det A) (oper 4) (machine 1) (time 8))
(route (det B) (oper 1) (machine 1) (time 8))
(route (det B) (oper 2) (machine 5) (time 4))
(route (det B) (oper 2) (machine 2) (time 6))
(route (det B) (oper 3) (machine 6) (time 3))
(route (det B) (oper 3) (machine 5) (time 2))
(route (det B) (oper 4) (machine 7) (time 2))
(route (det B) (oper 4) (machine 2) (time 3))
...
)
(deffacts machines
(machine (num 1))
(machine (num 2))
(machine (num 3))
(machine (num 4))
(machine (num 5))
(machine (num 6))
(machine (num 7))
(machine (num 8))
)
(deffunction my-predicate (?fact1 ?fact2)
(< (fact-slot-value ?fact1 time) (fact-slot-value ?fact2 time)))
(deffunction find-max2 (?template1 ?predicate1 ?operation ?template2 ?max)
(bind ?max FALSE)
(do-for-all-facts ((?f2 ?template2)) (eq (fact-slot-value ?f2 count) 0)
(do-for-all-facts ((?f1 ?template1)) (eq (fact-slot-value ?f1 oper) ?operation) (eq (fact-slot-value ?f1 oper)(fact-slot-value ?f2 oper))
(if (or (not ?max) (funcall ?predicate1 ?f1 ?max))
then
(bind ?max ?f1)))
)
)
;--------------------------------------------------------------------------------------------------
(defrule work_on_1
(declare (salience 10000))
(machine (num ?num1)(count ?count1) (time ?time1))
(test (eq ?count1 0))
(test (eq ?time1 0))
?node1 <- (machine (num ?num1)(count ?count1) (time ?time1))
(details (det ?detail) (oper ?operation1) (count ?count2))
(test (not (eq ?count2 0)))
?node2 <- (details (det ?detail) (oper ?operation1) (count ?count2))
; add this code
(funcall find-max2 route my-predicate ?operation1 details ?max)
(test (eq ?operation1(fact-slot-value ?max oper )))
(route (machine ?num1) (det ?detail) (oper ?operation2) (time ?time2))
(test (eq ?operation2 (+ ?operation1 1)))
=>
(if (> ?count2 30)
then
(modify ?node1 (count 30) (time ?time2) (oper ?operation2) (det ?detail))
(modify ?node2 (count (- ?count2 30)))
(printout t ?*time*" ," ?num1 " 30 деталей типа "?detail " , " ?operation2 " , " ?time2 crlf)
else
(modify ?node1 (count ?count2) (time ?time2) (oper ?operation2) (det ?detail))
(modify ?node2 (count (- ?count2 ?count2)))
(printout t ?*time*" , " ?num1 " " ?count2 " , "?detail " , " ?operation2 " , " ?time2 crlf)
)
)

Answer 1

There are numerous issues with the code. First it's unclear why there’s redundant pattern matching for the machine and details facts:
(machine (num ?num1)(count ?count1) (time ?time1))
(test (eq ?count1 0))
(test (eq ?time1 0))
?node1 <- (machine (num ?num1)(count ?count1) (time ?time1))
(details (det ?detail) (oper ?operation1) (count ?count2))
(test (not (eq ?count2 0)))
?node2 <- (details (det ?detail) (oper ?operation1) (count ?count2))
The following code is sufficient to achieve the same task:
?node1 <- (machine (num ?num1) (count ?count1&0) (time ?time1&0))
?node2 <- (details (det ?detail) (oper ?operation1) (count ?count2&~0))
Second, it looks like you're expecting the funcall pattern to invoke a function call and place the result of that function call in the variable ?max:
(funcall find-max2 route my-predicate ?operation1 details ?max)
There is no funcall conditional element that has this type of behavior. In this case, you've just created a pattern conditional element that's looking for an ordered fact with the relation name funcall. If you wanted to invoke funcall from the conditions of a rule you'd use the test conditional element:
(test (funcall find-max2 route my-predicate ?operation1 details))
Since the function being invoked is a literal, find-max2, there's really no point in using funcall since you can just call the function directly:
(test (find-max2 route my-predicate ?operation1 details))
Third, there's not much point to using the query functions within the conditions of a rule since you can directly pattern match on the facts instead. In addition, the placement of the query function you've used within the conditions will cause it to be executed once there are matching machine/details facts, but possibly before there are any route facts, so it will likely not even return the correct results.
The following rule shows how to find the maximum values without using query functions:
(defrule find-max
(route (oper ?oper1) (time ?time1))
(exists (details (count 0) (oper ?oper1)))
(not (and (route (oper ?oper2) (time ?time2&:(> ?time2 ?time1)))
(exists (details (count 0) (oper ?oper2)))))
=>
(printout t "Maximum time is " ?time1 crlf))

Answer 2

(defrule work_on_1
(declare (salience 10000))
?node1 <- (machinegun (num ?num1) (count ?count1&0) (time ?time1&0))
?node2 <- (store (det ?detail) (oper ?operation1) (count ?count2&~0))
(tex_route (oper ?oper1) (time ?time2))
(exists (store (count 0) (oper ?oper1)))
(not (and (tex_route (oper ?oper2) (time ?time3&:(> ?time3 ?time2)))
(exists (store (count 0) (oper ?oper2)))))
(test (eq ?time1 ?time2))
(route (machine ?num1) (det ?detail) (oper ?operation2) (time ?time2))
(test (eq ?operation2 (+ ?operation1 1)))
=>
(if (> ?count2 30)
then
(modify ?node1 (count 30) (time ?time2) (oper ?operation2) (det ?detail))
(modify ?node2 (count (- ?count2 30)))
(printout t ?*time*" ," ?num1 " 30 деталей типа "?detail " , " ?operation2 " , " ?time2 crlf)
else
(modify ?node1 (count ?count2) (time ?time2) (oper ?operation2) (det ?detail))
(modify ?node2 (count (- ?count2 ?count2)))
(printout t ?*time*" , " ?num1 " " ?count2 " , "?detail " , " ?operation2 " , " ?time2 crlf)
)
)
It is new rule, but when the expert system stopped working - simply displays "1".